Difference between revisions of "2002 AMC 8 Problems/Problem 18"
Mrdavid445 (talk | contribs) (Created page with "Gage skated 1 hr 15 min each day for 5 days and 1 hr 30 min each day for 3 days. How long would he have to skate the ninth day in order to average 85 minutes of skating each day ...") |
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<math> \text{(A)}\ \text{1 hr}\qquad\text{(B)}\ \text{1 hr 10 min}\qquad\text{(C)}\ \text{1 hr 20 min}\qquad\text{(D)}\ \text{1 hr 40 min}\qquad\text{(E)}\ \text{2 hr} </math> | <math> \text{(A)}\ \text{1 hr}\qquad\text{(B)}\ \text{1 hr 10 min}\qquad\text{(C)}\ \text{1 hr 20 min}\qquad\text{(D)}\ \text{1 hr 40 min}\qquad\text{(E)}\ \text{2 hr} </math> | ||
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| + | ==Solution== | ||
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| + | Converting into minutes and adding, we get that she skated <math>75*5 +90*3 +x=375 +270 +x=645 +x</math> minutes total, where x is the amount she skated on day 9. Dividing by 9 to get the average, we get <math>\frac{645+x}{9}=85</math>. Solving for x, <cmath>645+x=765</cmath> <cmath>x=120</cmath> Now we convert back into hour and minute to get 2 hours. <math>\text{(E)}</math> | ||
Revision as of 12:30, 5 July 2012
Gage skated 1 hr 15 min each day for 5 days and 1 hr 30 min each day for 3 days. How long would he have to skate the ninth day in order to average 85 minutes of skating each day for the entire time?
Solution
Converting into minutes and adding, we get that she skated 
minutes total, where x is the amount she skated on day 9. Dividing by 9 to get the average, we get 
. Solving for x, ![\[645+x=765\]](http://latex.artofproblemsolving.com/1/4/2/1427cf5f8cb037657e41bdb1630cc14a495e7c2b.png)
![\[x=120\]](http://latex.artofproblemsolving.com/b/d/9/bd9f7a95736540cfe5ac1d90c69c4c2801cbf11f.png)
Now we convert back into hour and minute to get 2 hours. 
